Tarjan-割点

割点——tarjan

---

 1 #include 
 2 using namespace std;
 3 
 4 const int MAXN = 20001;
 5 const int MAXM = 100001;
 6 int n, m;
 7 int ans;//个数 
 8 
 9 
10 int head[MAXN], cnt, v[2 * MAXM], nxt[2 * MAXM];
11 void add(int x, int y) {
12     nxt[++cnt] = head[x];
13     head[x] = cnt;
14     v[cnt] = y;
15 }
16 
17 
18 int dfn[MAXN], low[MAXN], ind;
19 int cut[MAXN];
20 
21 
22 void tarjan(int now, int fa) {
23     dfn[now] = low[now] = ++ind;
24     int child = 0;
25     for (int i = head[now]; i ; i = nxt[i]) {
26         if(!dfn[v[i]]) {
27             tarjan(v[i], now);
28             low[now] = min(low[now], low[v[i]]);
29             
30             if(low[v[i]] >= dfn[now] && now != fa) cut[now] = 1;//不为根节点
31             if(now == fa) child++; //为根节点，子树统计 
32         }
33         else {
34             low[now] = min(low[now], dfn[v[i]]);
35         }
36     }
37     if(now == fa && child >= 2) cut[now] = 1;//为根节点且有两个及以上的子节点 
38 }
39 int main() {
40     
41     scanf("%d%d", &n, &m);
42     for (int i = 1; i <= m; i++) {
43         int x, y;
44         scanf("%d%d", &x, &y);
45         add(x, y);
46         add(y, x); 
47     }
48     
49     for (int i = 1; i <= n; i++)
50         if(!dfn[i]) 
51             tarjan(i, i);
52     
53     for (int i = 1; i <= n; i++)
54         if(cut[i] == 1)
55             ans++;        
56     printf("%d\n", ans);
57     for (int i = 1; i <= n; i++)
58         if(cut[i] == 1)
59             printf("%d ", i);
60             
61             
62     return 0;
63 }